TypeError:“tuple”对象不支持在非tuple对象上分配项

Question

此代码采用一个有序的元组列表（最高分数到最低分数）并收集最高、第二高和第三高得分者的姓名和分数。如果他们打成平手，这两个名字都会加到同一个名单上。在

myresults=[('Raven', '18'), ('Cobra', '8'), ('Lion', '6'), ('Otter', '2')]


FirstScore=myresults[0][1]
SecondHighestScore=0
ThirdHighestScore=0
for i in myresults:
    if i[1]==FirstScore:
        FirstPlacePatrols.append(i[0])
for i in myresults:
    print(i[1])
    print(repr(i[1]))
    if int(i[1])<int(FirstScore):
        if int(i[1])>=SecondHighestScore:
            print(i[1])
            i[1]=SecondHighestScore
            SecondPlacePatrols.append(i[0])
for i in myresults:
    if int(i[1])<SecondHighestScore:
        if int(i[1])>=ThirdHighestScore:
            i[0]=ThirdHighestScore
            ThirdPlacePatrols.append(i[0])
print(FirstPlacePatrols)
print(FirstScore)
print(SecondPlacePatrols)
print(SecondHighestScore)
print(ThirdPlacePatrols)
print(ThirdHighestScore)

但是

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收益率

TypeError: 'tuple' object does not support item assignment

尽管如此

print(repr(i[1]))

屈服

'18'

显然不是元组。在

Answer 1

您不能更改tuple()s-它们是不可变的。你可以创造一个新的。或者，您可以使用itertools.groupby将元组组合在一起并执行一些选择性输出：

myresults=[('Raven', '18'), ('Cobra', '8'), ('Lion', '6'), ('Swine', '6'), ('Otter', '2')]

from itertools import groupby

grped = groupby(myresults, lambda x: int(x[1])) 

# create a dict for all results
result_dict = {}
for key in grped :
    result_dict[key[0]] = [value for value,_ in key[1]] 

# print top 3 results:
for k in sorted(result_dict,reverse=True):
    print(k)
    print(result_dict[k])

# whole dict
print(result_dict)

输出：

^{pr2}$

---

第二种解决方法是使用collections.defaultdict：

myresults=[('Raven', '18'), ('Cobra', '8'), ('Lion', '6'), ('Swine', '6'), ('Otter', '2')]

from collections import defaultdict

result_dict = defaultdict(list)

for value,key in myresults:
    result_dict[int(key)].append(value)

for k in sorted(result_dict,reverse=True):
    print(k)
    print(result_dict[k])

print(result_dict)


18
['Raven']
8
['Cobra']
6
['Lion', 'Swine']
2
['Otter']

# whole dict
defaultdict(<class 'list'>, {18: ['Raven'], 8: ['Cobra'], 
                              6: ['Lion', 'Swine'], 2: ['Otter']})

多库语：

• itertools.groupby()
• collections.defaultdict()
• sorted(iterable[,key=...])

Answer 2

我的解决方案是：

from collections import defaultdict
given_list = [('Raven', '18'), ('Cobra', '8'), ('Lion', '6'), ('Python', '6'),('Otter', '2')]
reversed_dict = defaultdict(list)
for key,value in given_list:
    reversed_dict[int(value)].append(key)

for k in reversed(sorted(reversed_dict)[-3:]):
     print(k,reversed_dict[k])

输出：

^{pr2}$